What is the area of the region between the graphs of $f(x)=3x^2+3x-8$ and $g(x)=2x^2+x+7$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{160}{3}$ (Choice B) B $\dfrac{256}{3}$ (Choice C) C $34$ (Choice D) D $91$
Solution: Visualizing the area We sketch the graphs of $f$ and $g$ first. ${2}$ ${\llap{-}2}$ ${\llap{-}4}$ ${20}$ ${40}$ $f$ $g$ $y$ $x$ From the graph, it appears that $g(x)\ge f(x)$ between the points where the graphs intersect. From this we are looking to evaluate: $ \int_{a}^{b}\left( g(x)-f(x) \right)\,dx$ where $a$ and $b$ are the $x$ -coordinates of the points of intersection. Finding the $x$ -coordinates of the intersection points We can find the $x$ -coordinate of each point of intersection by setting the functions equal to each other and solving the resulting equation. $\begin{aligned} f(x)&=g(x) \\\\ 3x^2+3x-8&=2x^2+x+7\\\\ x^2+2x-15 &= 0\\\\ (x+5)(x-3)&= 0 \end{aligned}$ The graphs intersect where $x=-5$ and $x=3$. Setting up the definite integral Thus, the area of the shaded region pictured above is given by: $\begin{aligned} &\phantom{=} \int_{-5}^{3}\left(2x^2+x+7-(3x^2+3x-8)\right)\,dx \\\\ &= \int_{-5}^{3}\left(-x^2-2x+15\right)\,dx \end{aligned}$ Evaluating the definite integral $\begin{aligned} &\phantom{=} \int_{-5}^{3}\left(-x^2-2x+15\right) \,dx \\\\ &= -\dfrac{x^3}{3}-x^2+15x~\Bigg|_{-5}^{3} \\\\ &= \left( -9-9+45 \right) - \left( \dfrac{125}{3} -25 -75\right) \\\\ &= \dfrac{256}{3} \end{aligned}$ Answer The area is $\dfrac{256}{3}$ square units.